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location Germany
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visits member for 1 year, 8 months
seen Mar 19 at 14:09

Jul
31
awarded  Teacher
Jul
25
comment Why is every Divisor of the rational function field $K(x)$ over $K$ a principal divisor if $K$ is algebraically closed
So the statement is wrong as I suggested. Because the degree of a prime divisor is 1.
Jul
24
asked Why is every Divisor of the rational function field $K(x)$ over $K$ a principal divisor if $K$ is algebraically closed
Jun
5
comment What is meant by $|dxdy|^{1/2}$ in the integral?
What does the symbol $\wedge$ mean and thus $dx \wedge dy$?
Jun
5
comment What is meant by $|dxdy|^{1/2}$ in the integral?
And how would this expression change under changes of coordinates? I'm sorry if this is fundamental, but I'm not very familiar with analysis this deep. We had the projective coordinates: $x = x$ and $s = y/x$. So how would the written object change under the change of coordinates to those above?
Jun
5
asked What is meant by $|dxdy|^{1/2}$ in the integral?
Apr
16
comment Characterization of transcendental elements in algebraic function fields
@ YACP: Thank you very much for your help!
Apr
16
accepted Characterization of transcendental elements in algebraic function fields
Apr
16
comment Characterization of transcendental elements in algebraic function fields
Now I got it. If $z$ is transcendental over $K(X)$, then $K(z) \cong K(X)$ and $X$ is clearly algebraic over $K(X)$ since $X\cdot X^1 + X^2 \cdot X^0 = 0$ and $X \neq 0$ and $X^2 \neq 0$.
Apr
16
comment Characterization of transcendental elements in algebraic function fields
To the second part: Why is $z$ algebraic over $K(X)$, if $z$ is transcendental over $K$? EDIT: I haven't met him yet.
Apr
15
comment Characterization of transcendental elements in algebraic function fields
Does this change the situation to a convincing one? And yes, I'm new to field theory.
Apr
15
awarded  Editor
Apr
15
revised Characterization of transcendental elements in algebraic function fields
added 178 characters in body
Apr
15
asked Characterization of transcendental elements in algebraic function fields
Jan
22
answered Book/tutorial recommendations: acquiring math-oriented reading proficiency in German
Jan
19
comment Prove the Borel Lemma
It should have been: \begin{align} f^{(m)}(x) =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} (x^k \cdot \chi_k(x))^{(m)}\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( \sum\limits_{\beta=0}^{m}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right)\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( (x^k)^{(m)}\cdot \chi_k(x)+\sum\limits_{\beta=0}^{m-1}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right) \end{align}
Jan
19
comment Prove the Borel Lemma
Now we should ensure that $f$ is doing the trick: $f^{(k)}(0) = a_k$. Because of $\chi_k(0) = 1$ and $(x^k)^{(m)}\Big|_{x=0} \neq 0$ only for $m=k$ it follows that $$ f^{(k)}(0) = \frac{a_k}{k!} \left( k! \right) = a_k.$$ If there are no serious misleadings in my thoughts the only thing that misses is to find the proper $(r_k)$ ensuring that $\chi_k$ is cutting off right, even if $(a_k)$ is divergent. This $(r_k)$ should be directly related with $(a_k)$.
Jan
19
comment Prove the Borel Lemma
where $$(x^k)^{(m)} = \begin{cases} \frac{k!}{(m-1)!}\ x^{k-m}&,\ m<k\\ k!&,\ m=k\\ 0&,\ m>k \end{cases} $$ It should be $\chi_k(x) \leq c_k, c_k \in \mathbb{C}$ and $\chi_k^{(m)}(x)$ cutting off (for a proper $(r_k)$) the trouble making $(x^k)^{(\beta)}$ for $|x| >1$, hence $f \in C^{\infty}(\mathbb{C})$ even if $(a_k)$ is divergent, but only fo a proper series $(r_k)$ letting $\chi_k^{(m-\beta)}(x)$ cut off those $a_k$ making trouble. For $a_k$ convergent this is obvious.
Jan
19
comment Prove the Borel Lemma
Now I observe the $m-$th derivative: $$\begin{align} f^{(m)}(x) =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} (x^k \cdot \chi_k(x))^{(m)}\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( \sum\limits_{\beta=0}^{m}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right)\\ =& \sum\limits_{k=0}^{\infty} \frac{a_k}{k!} \left( (x^k)^{(m)}\cdot \chi_k(x)+\sum\limits_{\beta=1}^{m}{m \choose \beta}\ (x^k)^{(\beta)} \cdot \chi_k(x)^{(m-\beta)}\right) \end{align} $$
Jan
19
comment Prove the Borel Lemma
So then $$\chi_k^{(m)}(x) = \begin{cases} 0,& |x| \leq r_k \text{ or } |x| \geq r_k +1\\ \in C^{\infty},& \text{ else} \end{cases} .$$